ГДЗ по алгебре 9 класс Макарычев ФГОС Задание 267

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\[\textbf{а)}\ x² < 16\]

\[x^{2} - 16 < 0\]

\[(x - 4)(x + 4) < 0\]

\[x \in ( - 4;4).\]

\[\textbf{б)}\ x² \geq 3\]

\[x^{2} - 3 \geq 0\]

\[\left( x - \sqrt{3} \right)\left( x + \sqrt{3} \right) \geq 0\]

\[x \in \left( - \infty;\ - \sqrt{3} \right\rbrack \cup \left\lbrack \sqrt{3}; + \infty \right).\]

\[\textbf{в)}\ 0,2x² > 1,8\]

\[0,2x^{2} - 1,8 > 0\]

\[x^{2} - 9 > 0\]

\[(x - 3)(x + 3) > 0\]

\[x \in ( - \infty;\ - 3) \cup (3;\ + \infty).\]

\[\textbf{г)} - 5x^{2} \leq x\]

\[5x^{2} + x \geq 0\]

\[x(5x + 1) \geq 0\]

\[x \in ( - \infty; - 0,2\rbrack \cup \lbrack 0; + \infty).\]

\[\textbf{д)}\ 3x² < - 2x\]

\[3x^{2} + 2x < 0\]

\[x(3x + 2) < 0\]

\[x \in \left( - \frac{2}{3};0 \right).\]

\[\textbf{е)}\ 7x < x²\]

\[x^{2} - 7x > 0\]

\[x(x - 7) > 0\]

\[x \in ( - \infty;0) \cup (7; + \infty).\]